Question: What is the extraneous solution to these equations? $\dfrac{x^2 - 2x}{x + 10} = \dfrac{-20x - 80}{x + 10}$
Multiply both sides by $x + 10$ $ \dfrac{x^2 - 2x}{x + 10} (x + 10) = \dfrac{-20x - 80}{x + 10} (x + 10)$ $ x^2 - 2x = -20x - 80$ Subtract $-20x - 80$ from both sides: $ x^2 - 2x - (-20x - 80) = -20x - 80 - (-20x - 80)$ $ x^2 - 2x + 20x + 80 = 0$ $ x^2 + 18x + 80 = 0$ Factor the expression: $ (x + 8)(x + 10) = 0$ Therefore $x = -8$ or $x = -10$ At $x = -10$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -10$, it is an extraneous solution.